PHP - Moved: Difficult Where Clause

I notice plenty of CakePHP plugins don't have composer.json files in their github repositories (I'm not sure where composer downloads from if you don't specify a repository, is it github?) so I can't automatically download and update them. For composer to recognise and download the plugins, all thats needed is the composer.js file itself isn't it? Why wouldn't plugin authors just spend an extra 10 minutes and make a composer.js file? I'm guessing theres something I'm missing here, it must not be that simple. I only started using composer so I don't really know what its about yet.

Hi There!

I am quite far with this script I'm creating, but there is one thing I am not getting straight.

The script shows and hides information from a database by changing checkboxes. This can be filtered by a form.

Now my final problem: the checkboxes should start CHECKED instead of unchecked as they start now. It could be done by giving the $film or the $a attribute a start value I guess, but I don't know how.

I tried to put
$a = "checked=\"checked\"";
$film_query = "film='1'";
on several places in the beginning of the script, but it doesn't seem to affect anything. It only works if I put it right before the form, but then it checks the checkboxes every time after the form has been submitted (making it impossible to uncheck it). Not good

Could you give me a little help maybe?

Thanks a LOT!

Hi.  I am building an update table form.  In this table there are only two fields, Header and Intro.  Now I have the form and currently it displays whats already in the database.  Now when the user edits these two fields and presses edit.  Obviously I want it to update these fields.  I have the code here but there is an issue with it.  At the end of the update I believe I need a update fields WHERE.......  which would be used if there was more than one entrance in the table but there is and only ever will be one entry.

WHERE what?  I dont have a where anything.  Do I need this or can it be omitted somehow?

<?php
if(isset($_POST['edit'])){ 
//Process data for validation 
$header     = trim($_POST['header']);    
$intro     = trim($_POST['intro']);        
//Perform validations next 

//Prepare data for db insertion        
$header = mysql_real_escape_string($header);    
$intro      = mysql_real_escape_string($intro);        
//insert
$qUpdateDetails = "UPDATE index SET 
`header` = '".$header."', 
`intro` = '".$intro."',
 
WHERE ???????='".?????????."'";

$rUpdateDetails = mysql_query($qUpdateDetails) or die(mysql_error()); 
//next page
$page_edit = "Index page has been edited";
$url = "signed.php?edit='".$page_edit."'" or die(mysql_error());        
header("Location: ".$url."") or die(mysql_error()); 
 exit();
}     
}
?>
 

Hi   I am now on to viewing listings but want to to show listings that are just submitted by that user and display them on their profile page but can't get the WHERE clause working   Below is the coding for what I have   Profile page

echo "<p><a href='view-private-listings.php?id={$_SESSION['user_id']}'>View Listings</a></p>";  

Add Listing page HTML

Submitted By: <input type="text" name="submittedby">

View Listings Page

<?php
ini_set('display_startup_errors',1);
ini_set('display_errors',1);
error_reporting(-1);
?>

<?php 
$title = "Private Seller Listings";

include ( 'includes/header.php' );
?>

<?php
require_once("functions.php");

$con=mysqli_connect("host","username","password","database");
// Check connection
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

// Start a session for error reporting
session_start();

$submittedby=$_POST['submittedby'];
$listingtitle=$_POST['listingtitle'];

$query = mysqli_query($con,"SELECT listingtitle FROM privatelistings WHERE item LIKE '%$submittedby%'");

echo "<p>Title: {$listingtitle['listingtitle']}</p>";

mysqli_close($con);

?>

<?php include( 'includes/footer.php' ); ?>

All I get is the following error   Notice: Undefined index: listingtitle in /zacs-car-site/view-private-listings.php on line 26   Thank you in advance   Kind regards   Ian

hi,
how can I add more than one ID to a WHERE clause?
Code: [Select]
"SELECT * FROM t_mytable WHERE id_product = 1 ";
I need to select products with different id's i.e
Code: [Select]
"SELECT * FROM t_mytable WHERE id_product = 1,2,3,4,5 ";
Thanks

This may be super simple, but it's stumping me. Is this an If/Then clause: Code: [Select]
($month != 1 ? $month - 1 : 12)
Is it saying if $month does not equal 1 then $month-1; if so, then what does the colon mean? If not, then what is this piece of code doing?

I am brand new to mysql and php but I have created a database and loaded two tables using cPanel and phpMyAdmin. Now I need some programs to access my data. I have a couple of simple ones that work, but I can't figure out what I really need,

I am trying to Select a table Where the Value is a $variable, not a fixed value.

Of course the end result will be to pass the value from a Form, but I have to get this to work first.




<?php

// Connect to database=============================
include("connect_db.php"); 
$table='airplanes';
$amano='123456'
$iden='1'

// Send query ===========================================================
// $result = mysql_query("SELECT * FROM {$table} where ama='123456'"); == this works
// $result = mysql_query("SELECT * FROM {$table} where ama='940276'"); == this works
// $result = mysql_query("SELECT * FROM {$table} where id='1'"); // this works
// $result = mysql_query("SELECT * FROM {$table} where id = '{$iden}'"); == doesnt work
// $result = mysql_query("SELECT * FROM {$table} where id = $iden"); == doesnt work
// $result = mysql_query("SELECT * FROM {$table} where id = ($iden)"); // == doesnt work
// $result = mysql_query("SELECT * FROM {$table} where id = $iden"); // == doesnt work
// $result = mysql_query("SELECT * FROM {$table} where ama='$amano'"); // == doesnt work
$result = mysql_query("SELECT * FROM {$table} where ama=($amano)"); // == doesnt work


Thanks

Code: [Select]
$search = "`title` LIKE '%Silvin AND wts%'";
I want to search for both of those keywords why not work?

Hi,

I have a SQL statement:
$sql='SELECT PROPID, ADDRESS_1, ADDRESS_2, TOWN, POSTCODE1, POSTCODE2, BEDROOMS, BATHROOMS, RECEPTIONS, PRICE, TRANS_TYPE_ID FROM properties WHERE TRANS_TYPE_ID=2 AND BEDROOMS =3 AND PROP_SUB_ID IN (1,2,3,4,5,6,21,22,23,24,27,30,95,128,131) ORDER BY DATE_ADDED, PROPID DESC';

I want to be able to get the fields and their corresponding values as such:

$result['TRANS_TYPE_ID']=2;
$result['BEDROOMS']=3;

And then capture the values in PROB_SUB_IN IN(*)

How can I do this?

Any help will be greatly appreciated. Thanks

Hi,

I have an empty array and i populated the array through code and i printed to check if everything ok. So far so good. I now have an array say $ids = ('123','456'). Now, i have a table in db that has ids as well, I need to make a query to check for id's that are not inside the array, so if the db table have '123','456','789' i want the result of the query to be only the last one '789'.

$arr = array();
array_push($arr,$previd2);
print_r ($arr);

I get the result as follows:

Array ( [0] => 533349 [1] => 533355 )

so what query i can use to get the other id's from db table that is not part of that array?

here is what i tried:

$qry = "select staffname, joindate from staff WHERE OracleID NOT IN ($arr)";
$answ = mysqli_query($con, $qry);

I get error:

Notice
: Array to string conversion in
C:\xampp\htdocs\AttendanceSystem\login\reportsforallid.php

 

Edited March 30, 2020 by ramiwahdan
more info

hello.

i have a column called "pageName" and on the each page i have put $pName = "the name of that page".

i want to pull from the database all the information where the $pName on the page is the same as the pageName in the database

its getting the $pName = "adminHome" but it thinks its the name of the column not the row.

this is the error im getting:

Quote

DATABASE.PHP - confirm_query = MySQL Datatbase Query Failed: Unknown column 'adminHome' in 'where clause'

Last SQL query: SELECT * FROM pages WHERE pageName=adminHome



this is the function code
Code: [Select]
public static function find_by_pageName($pName=""){
global $database;
$sql = "SELECT * FROM ".self::$table_name." WHERE pageName=".$database->escape_value($pName)."";
$result_array = self::find_by_sql($sql);
return !empty($result_array) ? array_shift($result_array) : false;
}


this is the page code
Code: [Select]
<?PHP
require_once("../includes/initialize.php");

$pName = "adminHome";

$page = Pages::find_by_pageName($pName);
?>



thanks
rick

I have this code:


Code: [Select]
if (isset($_POST['update'])){
$ids = implode(",", array_map('intval', $_POST['m']));
$ranks = implode(",", array_map('intval', $_POST['ranks']));
if ($ids < 1)
message("Incorrect Data");
if ($ranks < 0)
message("Incorrect Data");
echo $ids;
echo "<br>";
echo $ranks;
exit;
$db->query('UPDATE friends set RANKS WHERE friend_id IN ('.$db->escape($ids).') AND user_id = '.$pun_user['id'].'') or error('Unable to remove users from online list', __FILE__, __LINE__, $db->error());
redirect("s.php?section=Friends","Thanks, Ranks Updated");
}
I grab the id's from the rank

This will spit out 3,4 for $ids and for $ranks, 1,2 as you can see, set for 1 and 2 respectivaly.

My problem is how can I use MYSQL to update the data correspond to each id using a IN clause?

Code: [Select]
$db->query('UPDATE friends set RANKS = "LOST $RANKS? HERE" WHERE friend_id IN ('.$db->escape($ids).') AND user_id = '.$pun_user['id'].'') or error('Unable to remove users from online list', __FILE__, __LINE__, $db->error());

Hey guys,

I don't know what to do. I've got a script used for gamble:


$gamble_rand = rand($number_of_tickets,$gamble_chance);
if($gamble_chance == $gamble_rand){


If $number_of_tickets is set by 99 and $gamble_chance is set by 100 the chance is 1:100 that the If-Clause is not true. But it seems to me that PHP is making its on "random". Independent of the values PHP is interpreting that 20 times of 100 times the correct logic is false...

If $number_of_tickets is set by 1 there's no big different. Its just a little bit less, but not the chance that it has to be.

Thanks for helping and Merry Christmas :-)

Hi,   Im having a problem and I can't seem to figure it out or find anything on the net.   If I use the following code the script successfully updates every row in the table:

mysqli_query($con,"UPDATE Ads SET Ads_LocalArea='Stroud'");

However if I try updating the table using the WHERE clause in any of the combinations below nothing happens.

mysqli_query($con,"UPDATE Ads SET Ads_LocalArea='Stroud' WHERE Ads_ID=$DBROWID");

----------------------------------------------------------------------   My Script:

mysqli_query($con,"INSERT INTO Ads (Ads_ID, Ads_AID, Ads_Title)
VALUES ('', '$Polished_AdRef', '$Polished_AdTitle')");

$DBROWID = mysqli_insert_id($con);

mysqli_query($con,"UPDATE Ads SET Ads_AID='Stroud' WHERE Ads_ID=$DBROWID");

// TRIED THESE TOO
// mysqli_query($con,"UPDATE Ads SET Ads_AID='Stroud' WHERE Ads_ID='$DBROWID'");
// mysqli_query($con,"UPDATE Ads SET Ads_AID='Stroud' WHERE Ads_ID='5'");

Does any one know where I am going wrong?

I had originally created a topic on my pagination not working, but I found out what the problem was. The topic I had created was irrelevant to what the problem was... but anyway, here's my situation:

I've got a query that searches in a MySQL table for a certain value. Then it list's all the findings in a table (pretty simple). Anyway, I obviously want to have a pagination system to make it easier to manage the amount of data per page.

The pagination works fine, it's used for another query that just pulls data (so NO WHERE clause is used!). That is what is causing the problem. I spent hours trying to figure out why this is happening, then I figured out that the only difference between the pagination that works and the one that doesn't is the query have the where clause.

What happens when the where clause is used is that the pagination works fine on the first page, but if you click the link to go to a different page, the page is empty and turns up no results so the mysql data that is pulled is lost.

So here is my code. I don't know exactly why the where clause causes the pagination to not work.

Code: [Select]
$conn = mysql_connect("$mysql_host","$mysql_user","$mysql_password") or trigger_error("SQL", E_USER_ERROR);
$db = mysql_select_db("$mysql_database",$conn) or trigger_error("SQL", E_USER_ERROR);

// find out how many rows are in the table
$sql = "SELECT COUNT(*) FROM ACTIVE WHERE MODEL='$model'";
$result = mysql_query($sql, $conn) or trigger_error("SQL", E_USER_ERROR);
$r = mysql_fetch_row($result);
$numrows = $r[0];

// number of rows to show per page
$rowsperpage = 5;
// find out total pages
$totalpages = ceil($numrows / $rowsperpage);

// get the current page or set a default
if (isset($_GET['currentpage']) && is_numeric($_GET['currentpage'])) {
   // cast var as int
   $currentpage = (int) $_GET['currentpage'];
} else {
   // default page num
   $currentpage = 1;
} // end if

// if current page is greater than total pages...
if ($currentpage > $totalpages) {
   // set current page to last page
   $currentpage = $totalpages;
} // end if
// if current page is less than first page...
if ($currentpage < 1) {
   // set current page to first page
   $currentpage = 1;
} // end if

// the offset of the list, based on current page
$offset = ($currentpage - 1) * $rowsperpage;


$sql = "SELECT * FROM ACTIVE WHERE MODEL='$model' ORDER BY INDEX_ID DESC LIMIT $offset, $rowsperpage";
$result = mysql_query($sql, $conn) or die('Error: ' . mysql_error());

echo "
<table width=100% align=center border=1 class=\"mod\"><tr>
 <td align=center bgcolor=#00FFFF><b>Rating</b></td>
 <td align=center bgcolor=#00FFFF><b>Title</b></td>
 <td align=center bgcolor=#00FFFF><b>Make/Model</b></td>
 <td align=center bgcolor=#00FFFF><b>Type</b></td>
 <td align=center bgcolor=#00FFFF><b>Difficulty</b></td>
 <td align=center bgcolor=#00FFFF><b>Comments</b></td>
 <td align=center bgcolor=#00FFFF><b>Views</b></td>
</tr>";
   
     while ($row = mysql_fetch_assoc($result)) {
     { // Begin while
      $title = $row["TITLE"];
      $type = $row["TYPE"];
      $difficulty = $row["DIFFICULTY"];
      $comments = $row["COMMENT_TOTAL"];
      $id = $row['INDEX_ID'];
      $rating = $row["RATING_AVERAGE"];
      $make = $row["MAKE"];
      $model = $row["MODEL"];
      $views = $row["HITS"];

     echo "
     <tr>
      <td>$rating/10</td>
      <td><a href=\"mod.php?id=$id\">$title</a></td>
      <td>$make - $model</td>
      <td>$type</td>
      <td>$difficulty</td>
      <td>$comments</td>
      <td>$views</td>
     </tr>";
    }
   }
   echo "</table><br />";
    /******  build the pagination links ******/
echo "<center>";
// range of num links to show
$range = 3;
echo "<i>Page: </i>";
// if not on page 1, don't show back links
if ($currentpage > 1) {
   // get previous page num
   $prevpage = $currentpage - 1;
   // show < link to go back to 1 page
   echo " <a stylehref='{$_SERVER['PHP_SELF']}?currentpage=$prevpage'>Previous</a> - ";
} // end if

// loop to show links to range of pages around current page
for ($x = ($currentpage - $range); $x < (($currentpage + $range) + 1); $x++) {
   // if it's a valid page number...
   if (($x > 0) && ($x <= $totalpages)) {
      // if we're on current page...
      if ($x == $currentpage) {
         // 'highlight' it but don't make a link
         echo " <b>$x</b> ";
      // if not current page...
      } else {
         // make it a link
         echo " <a href='{$_SERVER['PHP_SELF']}?currentpage=$x'>$x</a> ";
      } // end else
   } // end if
} // end for
                 
// if not on last page, show forward and last page links       
if ($currentpage != $totalpages) {
   // get next page
   $nextpage = $currentpage + 1;
    // echo forward link for next page
   echo " - <a href='{$_SERVER['PHP_SELF']}?currentpage=$nextpage'>Next</a> ";
} // end if
/****** end build pagination links ******/
mysql_close($conn);
}
echo "$pagination</center>";
?>

I really appreciate the help! Thanks.